$\int \frac{\sin x}{\sqrt{5 \sin ^2 x+6 \cos ^2 x}} \mathrm{~d} x=$

The integral $I = \int \frac{\sin x}{\sqrt{5\sin^2 x+6\cos^2 x}} dx$ simplifies by expressing the denominator in terms of $\cos^2 x$:

$5\sin^2 x + 6\cos^2 x = 5(\sin^2 x + \cos^2 x) + \cos^2 x = 5 + \cos^2 x$

Substituting $t = \cos x$ with $dt = -\sin x dx$ transforms the integral:

$I = -\int \frac{dt}{\sqrt{t^2 + 5}}$

Using the standard integral formula $\int \frac{dx}{\sqrt{x^2 + a^2}} = \log|x + \sqrt{x^2 + a^2}| + C$:

$I = -\log|t + \sqrt{t^2 + 5}| + C$

Reversing the substitution yields:

$I = -\log|\cos x + \sqrt{\cos^2 x + 5}| + C$

The expression inside the logarithm remains positive since $\sqrt{\cos^2 x + 5} > |\cos x|$ for all real $x$.

This result corresponds to option D.